Lecture 13: Sequence components (Tutorial)
1. Fig. 13.1 shows the single line diagram of a 13.8kV system connected to a 480V
bus through a 13.8k/480V transformer. Two motor loads of 400hp and 600hp are
connected to the bus through three parallel conductor copper cables. If a 3 phase
bolted fault occurs at F1 and F2 , compute the fault currents.
Ans: Let us take base power as 1000kVA and base voltage as 480V.
kVA 1000
Then base current 
3  base voltage
1000  1000

 1202.8 A
3  480
Base voltage ( per phase) 480 3

Base impedance 
Base current
1202.8
 0.2304
Now, we have to convert all important element into per unit values on a
common base. Here the important base is short circuit contribution from 13.8kV
source = 600MVA
600  10 6
Short circuit current 
 25102 A
3  13.8  10 3
X ratio = 15
R
base kVA
1000
Zs 

 0.00166 pu
short circuit kVA 600  10 3
Xs
 15 Z s  Rs2  X s2 Z s2  Rs2  X s2
Rs
i.e. (0.00166) 2  Rs2  (15 Rs ) 2
Rs  0.00011 pu X s  0.00165 pu
i.e. Z s in pu = 0.00011 + j0.00165
1000kVA Transformer.
ZT = 5.75% RT = 1.21%
%RT
base kVA
The per unit value of RT 

Transformer kVA 100
1000 1.21


 0.0121 pu
1000 100
1000 5.75

 0.0575 pu
Per unit value of ZT 
1000 100
XT 
Z T2  RT2  0.0562 pu
i.e. ZT in pu = 0.0121 + j0.0562
Cable C1
Length of cable C1  500m ,
Resistance of one conductor per km = 0.178 
Reactance of one conductor per km = 0.108 
Since, three conductors are in parallel, equivalent resistance and reactance for
500m length is given by,
0.178 500
Rc1 

 0.0297
3
1000
0.108 500
X c1 

 0.018
3
1000
Converting Rc1 and X c1 , into per unit,
Actual value 0.0297

Rc1 in pu 
base value
0.2304
 0.129
0.018
 0.078
X c1 in pu 
0.2304
i.e. Z c1 in pu = 0.129 + j0.078
Cable C2
Length of cable C2  300m
Resistance of one conductor per km is given as 0.181  and reactance / km is
given as 0.124  . Since, three conductors are in parallel, equivalent resistance
and reactance for 300m cable is given by,
0.181 300
Rc 2 

 0.0181
3
1000
0.124 300
X c2 

 0.0124
3
1000
Converting into pu
0.0181
 0.0786 pu
Rc 2 in pu 
0.2304
0.0124
 0.0538 pu
X c 2 in pu 
0.2304
i.e. Z c 2 in pu = 0.0786 + j0.0538
Motors
Assume that 1hp is equivalent to 1kVA
Subtranscient reactance = 25% X Ratio = 6
R
base kVA % X m1
Per unit reactance of motor 1 X m1 

motor kVA 100
1000 25


 0.625 pu
400 100
0.625
 0.1042 pu
6
For motor 2
1000 25
0.416
X m2 

 0.416 pu
Rm 2 
 0.069 pu
600 100
6
Z m 2 in pu = 0.069 + j0.416
Note that 1hp = 746w, if we assume a motor power factor of 0.746, then
equivalent motor MVA because 1kVA. Hence, the equivalent circuit of the
system is shown in Fig 13.2. The dotted lines indicate the ground potential.
Fault at F1 :
We now desire to compute Thevenin’s impedance at node A. For fault at F1 the
network as shown in fig 13.2 can be reduced to network as shown in fig 13.3.
Hence, total impedance, Z th is given by,
1
1
1
1



Z th Z s  ZT Z c1  Z m1 Z c 2  Z m 2
1
1
1



0.01221  j 0.05785 0.233  j 0.703 0.1476  j 0.4698
1
1
1



0.059 78.1 0.74 71.6 0.49 72.6
Rm1 
 16.95 78.1  1.35 71.6  2.04 72.6
 3.5  j16.6  0.426  j1.28  0.61  j1.9
 4.536  j19.78  20.3 77.1
1
Z th 
 0.049 77.1 pu
20.3 77.1
Therefore, three phase fault current at fault F1 
ba se current
Z th
1202.8
0.049
 24547A

Fault at F2
For fault at F2, the network shown in fig 2. can be reduced as shown in fig 4.
Calculation of Z
Z  Z c1 
1
1
1

Z s  ZT Z c 2  Z m 2
 0.129  j 0.078 
1
1
1

0.01221  j 0.05785 0.1476  j 0.4698
1
3.5  j16.6  0.61  j1.9
1
 0.129  j 0.078 
 0.129  j 0.078  0.054 77.5
4.11  j18.5
 0.129  j 0.078  0.012  j 0.053  0.141  j 0.131 pu
1
1
1
1
1
 


Z th Z Z m1 0.141  j 0.131 0.104  j 0.625
1
1


 3.83  j 3.53  0.26  j1.56
0.192 42.7 0.633 80.5
 0.129  j 0.078 
 4.09  j5.09  6.53 51.2
1
 0.153 51.2
i.e., Z th 
6.53 51.2
Therefore, the total three phase fault current at F2 

ba se ampere
per unit Z th
1202.8
 7861.44 A
0.126
Fig 13.5 shows the single line diagram of a 3 bus system. The sequence data for
transmission lines and generators are given in table 1. If a bolted single line to
ground fault occurs at F1, calculate the fault current. If the fault impedance is
j0.1pu; what will be the fault current?
Let us take E as 1 pu. For a SLG fault,
3E
Fault current 
where Z0 = Zero sequence impedance
Z0  Z1  Z 2  3Z f
Z1 = Positive sequence impedance
Z2 = Negative sequence impedance
We hav to find out the Thevenin’s equivalent zero, positive and negative
sequence impedances with respect to fault F.
Zero Sequence Impedance
For calculating Z0, the circuit shown in fig 13.5 is reduced as shown in fig 13.6.
i.e. j0.095 and j0.046 in parallel.
i.e. Z0  j 0.021  j 0.031  j 0.052 pu
Similarly, positive sequence impedance Z1 can be found out by reducing the
circuit as shown in fig 13.7.
i.e. Z1, positive sequence impedance = j0.01 + j0.124
= j0.134 pu
Negative sequence impedance Z2
For negative sequence impedance the circuit can be as shown in fig 13.8.
i.e. negative sequence impedance Z 2  j 0.01  j 0.08
= j0.09 pu
3E
3 1 0 pu

Z1  Z 2  Z 0 j 0.134  j 0.09  j 0.052
3 1 0
3 1 0


 10.869 90
j 0.276 j 0.276 90
If fault impedance Z f  j 0.1 , then
Now, fault current I f 
If 
3E
3 1 0 pu

 5.208 90
Z1  Z 2  Z0  3Z f
j 0.276  j 0.1 3